Question

A particle of mass $1kg$ is moving with velocity $\sqrt{2}m/s$ along a line $y=x+5$. Find the magnitude of angular momentum $inkg{m}^2/s$ of the particle about a perpendicular axis passing through origin $O$.

Answer 1

Given: $m=1kg$,$v=\sqrt{2}m/s$
Line of motion $y=x+5$
Compare given equation of the line of motion with
$y=mx+c$
we have $m=1$ or $\theta ={45}^{\circ }$ and $c=5\text{unit}$.
It represents a particle moving as shown in figure
Formula used: $L_O=mv{r}_{\perp }$
The linear momentum of the particle
$p=mv$
The angular momentum about O is given by
$L_O=mv{r}_{\perp }$
From figure $r_{\perp }=5\cos {45}^{\circ }=\frac{5}{\sqrt{2}}$
$L_O=\frac{5}{\sqrt{2}}mv$
Its direction is along negative $z-\text{axis}$.
Thus
$\overrightarrow{L}=-\frac{5}{\sqrt{2}}mv\hat{k}=-\frac{5}{\sqrt{2}}\times 1\times \sqrt{2}\hat{k}$
=$-5kgm/s^2\hat{k}$
FINAL ANSWER: $5$